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otez555 [7]
3 years ago
13

A volume of 80.0 mL of aqueous potassium hydroxide (KOH) is titrated against a standard solution of sulfuric acid (H2SO4). What

was the molarity of the KOH solution if 12.7 mL of 1.50 M H2SO4 was needed
Chemistry
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

0.478 M

Explanation:

Let's consider the neutralization reaction between KOH and H₂SO₄.

2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:

0.0127 L × 1.50 mol/L = 0.0191 mol

The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol

0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:

M = 0.0382 mol/0.0800 L = 0.478 M

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Answer:

A.  The entropy of the universe is increasing.(always)

Explanation:

The Second Law of Thermodynamics states that Entropy cannot decrease, because it keeps increasing and increasing and increasing. It will always stay on the increasing side.

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How to do balance chemical equiations
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3 years ago
How many moles of magnesium hydroxide, Mg(OH)2 can be created using 2.23 x 10^24 oxygen atoms?
igor_vitrenko [27]

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32.07 g/mole.

Explanation:

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7 0
3 years ago
What is the molarity of 2.00 L of a solution that contains 14.6 g NaCl?
White raven [17]

Answer: The molarity of the solution is 0.125 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of NaCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{14.6g}{58.5g/mol}=0.250mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.250mol}{2.00L}=0.125M

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8 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
3 years ago
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